How many numbers of 7 digits greater than 3500000 can be formed using the digits 1, 2, 2, 3, 3, 3, 7?

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My approach to solve this questions has been dividing the problem in 2 parts.



First Part.



The numbers that started with 37, since 22, 33, 31, etc. are not really options. So in the first two places I only have 1 possibility (37), leaving 1,2,2,3,3,7 available. So the numbers of combinations for this are:



$1times5times4times3times2times1=240$



Second Part.



The numbers that started with 7, leaving 1,2,2,3,3,3 available. So, the number of combinations for this are:



$1times6times5times4times3times2times1=720$



Therefore the numbers than can be formed are: $240+720=960$



Any help is greatly appreciated.




combinatorics discrete-mathematics permutations




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  • 2




    in the first part, you must have $1,2,2,3,3$.
    – farruhota
    Aug 22 at 4:56










  • Nice start, but the way you are counting, rearranging the two $2$'s would count differently in your total.
    – Cheerful Parsnip
    Aug 22 at 4:57










  • how would you count for $1,2,2$? by your method: $3cdot 2cdot 1=6$, whereas it must be $frac3!2!=3$.
    – farruhota
    Aug 22 at 5:00














up vote
0
down vote

favorite












My approach to solve this questions has been dividing the problem in 2 parts.



First Part.



The numbers that started with 37, since 22, 33, 31, etc. are not really options. So in the first two places I only have 1 possibility (37), leaving 1,2,2,3,3,7 available. So the numbers of combinations for this are:



$1times5times4times3times2times1=240$



Second Part.



The numbers that started with 7, leaving 1,2,2,3,3,3 available. So, the number of combinations for this are:



$1times6times5times4times3times2times1=720$



Therefore the numbers than can be formed are: $240+720=960$



Any help is greatly appreciated.




combinatorics discrete-mathematics permutations




share|cite|improve this question


















  • 2




    in the first part, you must have $1,2,2,3,3$.
    – farruhota
    Aug 22 at 4:56










  • Nice start, but the way you are counting, rearranging the two $2$'s would count differently in your total.
    – Cheerful Parsnip
    Aug 22 at 4:57










  • how would you count for $1,2,2$? by your method: $3cdot 2cdot 1=6$, whereas it must be $frac3!2!=3$.
    – farruhota
    Aug 22 at 5:00












up vote
0
down vote

favorite









up vote
0
down vote

favorite











My approach to solve this questions has been dividing the problem in 2 parts.



First Part.



The numbers that started with 37, since 22, 33, 31, etc. are not really options. So in the first two places I only have 1 possibility (37), leaving 1,2,2,3,3,7 available. So the numbers of combinations for this are:



$1times5times4times3times2times1=240$



Second Part.



The numbers that started with 7, leaving 1,2,2,3,3,3 available. So, the number of combinations for this are:



$1times6times5times4times3times2times1=720$



Therefore the numbers than can be formed are: $240+720=960$



Any help is greatly appreciated.




combinatorics discrete-mathematics permutations




share|cite|improve this question














My approach to solve this questions has been dividing the problem in 2 parts.



First Part.



The numbers that started with 37, since 22, 33, 31, etc. are not really options. So in the first two places I only have 1 possibility (37), leaving 1,2,2,3,3,7 available. So the numbers of combinations for this are:



$1times5times4times3times2times1=240$



Second Part.



The numbers that started with 7, leaving 1,2,2,3,3,3 available. So, the number of combinations for this are:



$1times6times5times4times3times2times1=720$



Therefore the numbers than can be formed are: $240+720=960$



Any help is greatly appreciated.





combinatorics discrete-mathematics permutations





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 22 at 7:18









N. F. Taussig

38.8k93153




38.8k93153










asked Aug 22 at 4:47









Grouper

658




658







  • 2




    in the first part, you must have $1,2,2,3,3$.
    – farruhota
    Aug 22 at 4:56










  • Nice start, but the way you are counting, rearranging the two $2$'s would count differently in your total.
    – Cheerful Parsnip
    Aug 22 at 4:57










  • how would you count for $1,2,2$? by your method: $3cdot 2cdot 1=6$, whereas it must be $frac3!2!=3$.
    – farruhota
    Aug 22 at 5:00












  • 2




    in the first part, you must have $1,2,2,3,3$.
    – farruhota
    Aug 22 at 4:56










  • Nice start, but the way you are counting, rearranging the two $2$'s would count differently in your total.
    – Cheerful Parsnip
    Aug 22 at 4:57










  • how would you count for $1,2,2$? by your method: $3cdot 2cdot 1=6$, whereas it must be $frac3!2!=3$.
    – farruhota
    Aug 22 at 5:00







2




2




in the first part, you must have $1,2,2,3,3$.
– farruhota
Aug 22 at 4:56




in the first part, you must have $1,2,2,3,3$.
– farruhota
Aug 22 at 4:56












Nice start, but the way you are counting, rearranging the two $2$'s would count differently in your total.
– Cheerful Parsnip
Aug 22 at 4:57




Nice start, but the way you are counting, rearranging the two $2$'s would count differently in your total.
– Cheerful Parsnip
Aug 22 at 4:57












how would you count for $1,2,2$? by your method: $3cdot 2cdot 1=6$, whereas it must be $frac3!2!=3$.
– farruhota
Aug 22 at 5:00




how would you count for $1,2,2$? by your method: $3cdot 2cdot 1=6$, whereas it must be $frac3!2!=3$.
– farruhota
Aug 22 at 5:00










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










You have to get rid of multiplicity as some digits are not distinguishable.



For the first case, choosing from $1,2,2,3,3$ should be



$$frac5!2!2!$$



For the second case, choosing from $1,2,2,3,3,3$ should be



$$frac6!2!3!$$



$$frac5!2!2!+frac6!2!3!=frac5!2!2!left(1+ frac63right)=frac3604=90$$






share|cite|improve this answer




















  • I was thinking about that, thank you for your answer and confirmation.
    – Grouper
    Aug 22 at 14:37










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
3
down vote



accepted










You have to get rid of multiplicity as some digits are not distinguishable.



For the first case, choosing from $1,2,2,3,3$ should be



$$frac5!2!2!$$



For the second case, choosing from $1,2,2,3,3,3$ should be



$$frac6!2!3!$$



$$frac5!2!2!+frac6!2!3!=frac5!2!2!left(1+ frac63right)=frac3604=90$$






share|cite|improve this answer




















  • I was thinking about that, thank you for your answer and confirmation.
    – Grouper
    Aug 22 at 14:37














up vote
3
down vote



accepted










You have to get rid of multiplicity as some digits are not distinguishable.



For the first case, choosing from $1,2,2,3,3$ should be



$$frac5!2!2!$$



For the second case, choosing from $1,2,2,3,3,3$ should be



$$frac6!2!3!$$



$$frac5!2!2!+frac6!2!3!=frac5!2!2!left(1+ frac63right)=frac3604=90$$






share|cite|improve this answer




















  • I was thinking about that, thank you for your answer and confirmation.
    – Grouper
    Aug 22 at 14:37












up vote
3
down vote



accepted







up vote
3
down vote



accepted






You have to get rid of multiplicity as some digits are not distinguishable.



For the first case, choosing from $1,2,2,3,3$ should be



$$frac5!2!2!$$



For the second case, choosing from $1,2,2,3,3,3$ should be



$$frac6!2!3!$$



$$frac5!2!2!+frac6!2!3!=frac5!2!2!left(1+ frac63right)=frac3604=90$$






share|cite|improve this answer












You have to get rid of multiplicity as some digits are not distinguishable.



For the first case, choosing from $1,2,2,3,3$ should be



$$frac5!2!2!$$



For the second case, choosing from $1,2,2,3,3,3$ should be



$$frac6!2!3!$$



$$frac5!2!2!+frac6!2!3!=frac5!2!2!left(1+ frac63right)=frac3604=90$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 22 at 4:58









Siong Thye Goh

80.4k1453101




80.4k1453101











  • I was thinking about that, thank you for your answer and confirmation.
    – Grouper
    Aug 22 at 14:37
















  • I was thinking about that, thank you for your answer and confirmation.
    – Grouper
    Aug 22 at 14:37















I was thinking about that, thank you for your answer and confirmation.
– Grouper
Aug 22 at 14:37




I was thinking about that, thank you for your answer and confirmation.
– Grouper
Aug 22 at 14:37












 

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